Arbitrariness of an arbitrary function after an operationConnecting functional identity of a function with...
In the world of The Matrix, what is "popping"?
Where do you go through passport control when transiting through another Schengen airport on your way out of the Schengen area?
PTIJ: Aliyot for the deceased
Why is my explanation wrong?
Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?
Was it really inappropriate to write a pull request for the company I interviewed with?
Does the in-code argument passing conventions used on PDP-11's have a name?
If nine coins are tossed, what is the probability that the number of heads is even?
Does the US political system, in principle, allow for a no-party system?
Why would the IRS ask for birth certificates or even audit a small tax return?
Is there a math expression equivalent to the conditional ternary operator?
The past tense for the quoting particle って
Is it a Cyclops number? "Nobody" knows!
What is the meaning of option 'by' in TikZ Intersections
What does "rhumatis" mean?
Why aren't there more gauls like Obelix?
Ultrafilters as a double dual
Quitting employee has privileged access to critical information
Affine transformation of circular arc in 3D
What is the purpose of a disclaimer like "this is not legal advice"?
Problems with rounding giving too many digits
When to use the term transposed instead of modulation?
What does it mean when I add a new variable to my linear model and the R^2 stays the same?
Can inspiration allow the Rogue to make a Sneak Attack?
Arbitrariness of an arbitrary function after an operation
Connecting functional identity of a function with its image setConstructing an iterative “signature” functionIs any real-valued function in physics somehow continuous?Precision in $f : A to B$ notationTangents and roots to simultaneous equations.How do we formally “identify” objects using isomorphisms?Domain of function $y$Notation for function spaces“Recursive definitions” in Tao's Analysis Vol IIs there a better definition for a function?
$begingroup$
Let’s say there is an arbitrary function f(x). Now I define another function g(x) = O(f(x)), where O is a mathematical operation (addition, inverse, derivative...whatever). My question is, can g(x) still be regarded as an “arbitrary function”?
My first thought is a no, because g(x) has to satisfy the mathematical operation as stated (i.e. it has to meet certain conditions/restrictions). But later I remember seeing statements such as “an arbitrary REAL number” (i.e. “the number has to be real” is a condition to be met), and now I’m confused.
functions
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let’s say there is an arbitrary function f(x). Now I define another function g(x) = O(f(x)), where O is a mathematical operation (addition, inverse, derivative...whatever). My question is, can g(x) still be regarded as an “arbitrary function”?
My first thought is a no, because g(x) has to satisfy the mathematical operation as stated (i.e. it has to meet certain conditions/restrictions). But later I remember seeing statements such as “an arbitrary REAL number” (i.e. “the number has to be real” is a condition to be met), and now I’m confused.
functions
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago
add a comment |
$begingroup$
Let’s say there is an arbitrary function f(x). Now I define another function g(x) = O(f(x)), where O is a mathematical operation (addition, inverse, derivative...whatever). My question is, can g(x) still be regarded as an “arbitrary function”?
My first thought is a no, because g(x) has to satisfy the mathematical operation as stated (i.e. it has to meet certain conditions/restrictions). But later I remember seeing statements such as “an arbitrary REAL number” (i.e. “the number has to be real” is a condition to be met), and now I’m confused.
functions
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let’s say there is an arbitrary function f(x). Now I define another function g(x) = O(f(x)), where O is a mathematical operation (addition, inverse, derivative...whatever). My question is, can g(x) still be regarded as an “arbitrary function”?
My first thought is a no, because g(x) has to satisfy the mathematical operation as stated (i.e. it has to meet certain conditions/restrictions). But later I remember seeing statements such as “an arbitrary REAL number” (i.e. “the number has to be real” is a condition to be met), and now I’m confused.
functions
functions
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
Kim Leung
asked 17 hours ago
Kim LeungKim Leung
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
Kim LeungKim Leung
11
asked 17 hours ago
Kim LeungKim Leung
11
11
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kim Leung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago
add a comment |
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As hinted by Minus One-Twelfth, the range of $g$ must be contained in the range of $O$. But that is the only restriction. That is, for any function $h$ on some domain $A$ whose
range is contained in the range of $O$, there is a function $f$ from $A$ into the domain of $O$ such that $h(x) = O(f(x))$ for all $x in A$. Namely, for each $x in A$ we let $f(x)$ be some $y$ such that $O(y) = h(x)$.
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Kim Leung is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138855%2farbitrariness-of-an-arbitrary-function-after-an-operation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As hinted by Minus One-Twelfth, the range of $g$ must be contained in the range of $O$. But that is the only restriction. That is, for any function $h$ on some domain $A$ whose
range is contained in the range of $O$, there is a function $f$ from $A$ into the domain of $O$ such that $h(x) = O(f(x))$ for all $x in A$. Namely, for each $x in A$ we let $f(x)$ be some $y$ such that $O(y) = h(x)$.
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
As hinted by Minus One-Twelfth, the range of $g$ must be contained in the range of $O$. But that is the only restriction. That is, for any function $h$ on some domain $A$ whose
range is contained in the range of $O$, there is a function $f$ from $A$ into the domain of $O$ such that $h(x) = O(f(x))$ for all $x in A$. Namely, for each $x in A$ we let $f(x)$ be some $y$ such that $O(y) = h(x)$.
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
As hinted by Minus One-Twelfth, the range of $g$ must be contained in the range of $O$. But that is the only restriction. That is, for any function $h$ on some domain $A$ whose
range is contained in the range of $O$, there is a function $f$ from $A$ into the domain of $O$ such that $h(x) = O(f(x))$ for all $x in A$. Namely, for each $x in A$ we let $f(x)$ be some $y$ such that $O(y) = h(x)$.
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
$endgroup$
As hinted by Minus One-Twelfth, the range of $g$ must be contained in the range of $O$. But that is the only restriction. That is, for any function $h$ on some domain $A$ whose
range is contained in the range of $O$, there is a function $f$ from $A$ into the domain of $O$ such that $h(x) = O(f(x))$ for all $x in A$. Namely, for each $x in A$ we let $f(x)$ be some $y$ such that $O(y) = h(x)$.
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
answered 16 hours ago
Robert IsraelRobert Israel
326k23215469
326k23215469
add a comment |
add a comment |
Kim Leung is a new contributor. Be nice, and check out our Code of Conduct.
Kim Leung is a new contributor. Be nice, and check out our Code of Conduct.
Kim Leung is a new contributor. Be nice, and check out our Code of Conduct.
Kim Leung is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138855%2farbitrariness-of-an-arbitrary-function-after-an-operation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$g$ may not be completely arbitrary anymore. For example, if the "O" is the squaring operation, so $g(x) = f(x)^2$, then $g(x)$ is now always non-negative (assuming $f$ is real-valued).
$endgroup$
– Minus One-Twelfth
17 hours ago
$begingroup$
Thanks for the answers first. So does it mean that a function can still be regarded as “arbitrary” even if there are restrictions on it? Somehow this looks contradictory to me....
$endgroup$
– Kim Leung
13 hours ago